# Work done to pump water out of a conical tank into a window above it

Water is pumped from a conical tank of top radius 3 ft and a height of 5 ft to a window 10 ft above the tank. The tank is completely full of water. How much work is done?

This is what I have so far: $$\int_0^{15}\pi(\text{radius of slice})^2 \cdot (62.4) (15-h)\,dh$$ I am having a problem figuring out the radius of the slice

• Hint: chop the conical tank into infinitesimal cylinders. – user137731 Nov 10 '15 at 19:25
• This is what i have so far: integral [0, 15] pi(radius of slice)^2 * (62.4) (15-h) I am having a problem figuring out the radius of the slice. – Darth Nov 10 '15 at 20:02
• You might tell us poor Europeans where that $62.4$ comes from. (Also you might set the upper limit of integration to $5$ instead of $15$.) – TonyK Nov 10 '15 at 23:20

To draw the problem in a coordinate system, choose a positive $x$ axis in the downward direction since the motion is vertical. Take the window at $x = 0$, the top of the tank at $x = 10$ and the bottom of the tank at $x = 15$. The partition of water is therefore the closed interval $[10,15]$ on the $x$ axis.
Take the radius of the cone at $x_i$, the function is $9-3x_i/5$. Then take an element of volume as a circular disk having thickness $\Delta_i x$ and the radius of the cone at $x_i$, the volume is $\Delta_i V = \pi (9-3x_i/5)^2 \Delta_i x$. Then take $62.4$ as the weight of water in pounds per cubic foot, the force required to pump an element is $\Delta_i F = 62.4 \pi (9-3x_i/5)^2 \Delta_i x$.
If $\Delta_i x$ is close to $0$, then the distance through which an element moves is approximately $x_i$. Thus the work done pumping an element to the window is $\Delta_i W = 62.4 \pi x_i (9-3x_i/5)^2 \Delta_i x$ and the integral is:
$$W = 62.4 \pi \int_{10}^{15} \! x(9-3x/5)^2 \, \textrm{d}x$$